4
$\begingroup$

If two topological spaces are weak homotopy equivalent to each other, are their Cech cohomology groups the same?

$\endgroup$
1
  • 2
    $\begingroup$ The obvious and well known (I am sure :-) answer is NO (regardless of is or are :-). $\endgroup$ Feb 21, 2015 at 12:02

2 Answers 2

4
$\begingroup$

$$T = \left\{ \left( x, \sin \frac{1}{x} \right ) : x \in (0,1] \right\} \cup \{(0,y)\mid y\in[-1,1]\}$$

This has trivial homotopy groups in degrees $\ge1$ but according to Wikipedia nontrivial Čech cohomology in degree 1.

$\endgroup$
4
$\begingroup$

To give a more enlightening answer to the question:

Cech cohomology is not the same as singular cohomology. However it is on CW-complexes. But there is CW approximation for topological spaces and singular cohomology is a weak homotopy invariant, so Cech cohomology can't be.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .